Construction of real interpolation Banach space Let $(A,\| \cdot \|_A)$, $(B,\| \cdot \|_B)$ and $(E,\| \cdot \|_E)$ real Banach spaces such that $A\subseteq E$ and $B\subseteq E$ with continuous injections. Let $0 < \theta < 1$. For $x\in E$ and $C\geq 0$ consider the following property on the pair $(x,C)$:
$$
(P) \hspace{10mm} \forall \epsilon > 0, \,\exists (a,b)\in A\times B \text{ such that } x = a + b, \, \|a\|_A\leq C\epsilon^{-\theta} \text{ and } \|b\|_B \leq C\epsilon^{1-\theta}.
$$
Let $[A,B]_{\theta,\infty}$ the set of all $x\in E$ such that there exists $C\geq 0$ such that $(x,C)$ verifies the property (P). It is easy to prove that if $(x_1,C_1)$ and $(x_2,C_2)$ verifies (P), then $(x_1 + x_2,C_1+C_2)$ verifies (P) and that for every real $\alpha$, $(\alpha x_1, |\alpha|C_1)$ verifies (P), thus $[A,B]_{\theta,\infty}$ is a vector subspace of $E$.
For every $x\in [A,B]_{\theta,\infty}$ define
$$
\|x\|_{[A,B]_{\theta,\infty}} = \inf\{C\geq 0 : (x,C) \text{ verifies (P)}\}.
$$
Then $\| \cdot\|_{[A,B]_{\theta,\infty}}$ is a norm on $[A,B]_{\theta,\infty}$.
My question is: Is this construction equivalent to the construction of the interpolation space $[A,B]_{\theta,\infty,K}$ obtained with the $K$-method? 
I can prove the inequality
$$
\|x\|_{[A,B]_{\theta,\infty,K}}\leq 2\|x\|_{[A,B]_{\theta,\infty}},
$$
where the norm in the left is the one obtained with the $K$-method.
How can I prove that the norms $\| \cdot\|_{[A,B]_{\theta,\infty,K}}$ and $\| \cdot\|_{[A,B]_{\theta,\infty}}$ are equivalent without requiring the open mapping theorem?
Thank you in advance!
 A: \begin{align*}
\Vert x\Vert_{\lbrack A,B]_{\theta,\infty,K}}  & =\sup_{t>0}t^{-\theta
}K(x,t)\\
K(x,t)  & =\inf\{\Vert a\Vert_{A}+t\Vert b\Vert_{B}:\,a+b=x\}.
\end{align*}
By the definition of infimum, given $t>0$ and $\delta>0$ there exist $a\in A$, $b\in B$
such that $a+b=x$ and $$
\Vert a\Vert_{A}+t\Vert b\Vert_{B}\leq K(x,t)+\delta t^{\theta}%
$$
and so$$
t^{-\theta}\Vert a\Vert_{A}+t^{1-\theta}\Vert b\Vert
_{B}\leq t^{-\theta}(K(x,t)+\delta t^{\theta})\leq\Vert x\Vert_{\lbrack
A,B]_{\theta,\infty,K}}+\delta,
$$
which shows that 
\begin{align*}
t^{-\theta}\Vert a\Vert_{A}  & \leq\Vert x\Vert_{\lbrack A,B]_{\theta
,\infty,K}}+\delta=:C_{0}\\
t^{1-\theta}\Vert b\Vert_{B}  & \leq\Vert x\Vert_{\lbrack A,B]_{\theta
,\infty,K}}+\delta=:C_{0}.
\end{align*}
By replacing $t$ with $1/\epsilon$ we have that property P holds and $C_{0}$ is an admissible constant $C$ in the
definition of $\Vert x\Vert_{\lbrack A,B]_{\theta,\infty}}$. Thus,$$
\Vert x\Vert_{\lbrack A,B]_{\theta,\infty}}\leq C_0=\Vert x\Vert_{\lbrack
A,B]_{\theta,\infty,K}}+\delta
$$
for every $\delta$. Letting $\delta\rightarrow0^{+}$ gives the inequality.
